Codeforces Round #370 (Div. 2) C

Description

Memory is now interested in the de-evolution of objects, specifically triangles. He starts with a equilateral triangle of side length x, and he wishes to perform operations to obtain an EQU Ilateral triangle of side length y.

In a single second, he can modify the length of a single side of the current triangle such that it remains a non-degenerat E Triangle (triangle of positive area). At any moment of time, the length of each side should is integer.

What's the minimum number of seconds required for Memory to obtain the equilateral triangle of side length y ?

Input

The first and only line contains integers x and y (3≤ y < x ≤100 -the starting and ending equilateral triangle side lengths respectively.

Output

Print a single integer-the minimum number of seconds required for Memory to obtain the equilateral triangle of side Leng th y if he starts with the equilateral triangle of side length x.

Examples Input

6 3

Output

4

Input

8 5

Output

3

Input

22 4

Output

6

Note

In the first sample test, Memory starts with an equilateral triangle of side length 6 and wants one of side length 3. Denote a triangle with sides a, b, and C as (a, b, c ). Then, the Memory can do.

In the second sample test, Memory can does.

In the third sample test, Memory can does:

.

Test instructions: Turn a big equilateral triangle into a small equilateral triangle that takes a few steps

Solution: We consider to be more convenient upside down, first increase one side to the maximum, then add the second edge, in turn, loop to the big equilateral triangle

#include <bits/stdc++.h>using namespace Std;int main () {int n,m;    int a,b,c;    int POS;    int cot=0;    cin>>n>>m;    A=m,b=m,c=m; while (a!=n| | b!=n| |            C!=n) {if (a!=n) {int pos1;            cout<< "A" <<endl;               for (int i=1; i<=n; i++) {//cout<< "B" <<endl;                cout<<a<< "" <<b<< "" <<c<<endl; if ((a+i) +b>c&& (b+c) > (a+i) && (a+i) +c>b&& (a+i) <=n) {PO                   S1=i;                cot++;            }} a=a+pos1;           cot++;           cout<<a<<endl;         Break        cout<<a<< "" <<b<< "" <<c<<endl;      }//cout<<a<< "" <<b<< "" <<c<<endl;        Break            if (b!=n) {int pos2; for (int I=1; i<=n;                i++) {//cout<< "B" <<endl; if ((b+i) +a>c&& (a+c) > (b+i) && (b+i) +c>a&& (b+i) <=n) {//                    B=b+i;                    Pos2=i;                cot++;            }} B=b+pos2;          cot++;        cout<<a<< "" <<b<< "" <<c<<endl;       }//cout<<a<< "" <<b<< "" <<c<<endl;        Break            if (c!=n) {int pos3; for (int i=1; i<=n; i++) {if (c+i) +a>b&&b+ (c+i) >a&&b+a>c+i&&                    (c+i) <=n) {//cout<<i<<endl;                   Pos3=i;                   C=c+i;                cot++;            }} C=pos3+c;         cot++; cout<<a<< "" <<b<< "" &LT;&Lt;c<<endl;       }//break;    cout<<a<< "" <<b<< "<<c<<endl;*/} cout<<cot<<endl; return 0;}

　　

Codeforces Round #370 (Div. 2) C